D=number of dimes
q=number of quarters
let's count everything in cents
dimes are worth 10 cents
quarters are worth 25 cents
10d+25q=700
divide both sides by 5
2d+5q=140
he has 7 more quarters than dimes
q=7+d
subsitute 7+d for q in other equation
2d+5q=140
2d+5(7+d)=140
2d+35+5d=140
7d+35=140
minus 35 both sides
7d=105
divide both sides by 7
d=15
sub back
q=7+d
q=7+15
q=22
22 quarters and 15 dimes
= x·[2(x²- 4x + 4) + 3(x² + 6x + 9) - x² + 5x]
= x(2x² - 8x + 8 + 3x² + 18x + 27 - x² + 5x)
= x(4x² + 15x + 35)
Answer: HK
Step-by-step explanation:
Answer:
Here's your answer
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<h3>
Answer: 19</h3>
Explanation:
Let's break 110 down into its prime factors
110 = 11*10
110 = 11*2*5
110 = 2*5*11
We have three different prime factors that multiply to 110. However, the instructions say there are 4 integers that multiply to 110. To fix this, we can say
110 = 1*2*5*11
now we see that 1,2,5 and 11 multiply out to 110
They add to 1+2+5+11 = 3+16 = 19
